If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3

Lesson 5: Topic B: Lessons 10-12: Logarithm properties

# Using the logarithmic power rule

Sal rewrites log₅(x³) as 3log₅(x). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• How do you do log2 -1
• This is undefined, because you cannot raise power to 2 in order to get -1.
• I'd like to see a step by step solution to the following equation: 5^(3x+2)=3^(4x-1) solve for x
• log 5^(3x+2) = log3^(4x-1)
(3x+2)log5 = (4x-1)log3
3xlog5 + 2log5 = 4xlog3 - log3
3xlog5 - 4xlog3 = -2log5 - log3
x(3log5 - 4log3) = -(2log5+log3)

x = -log75 / (3log5 -4log3)
• what's log2(2x) - log2x^3 = 5 ?
• simple just follow these steps!
1)log2(2x)-log2(x^3)=5 write in the form that is logbX-logbY=logbX/Y;
then this will become log2(2x/ (x^3) )=5
2)log2(2/x^2)=5
3) 2^5=2/x^2 => 32=2/x^2
= 32x^2=2
= x^2=1/16
10Q
• What happens if you have logarithms involving imaginary or complex numbers?
• Since you can't use normal methods to evaluate the log, you would need to resort to things like Euler's Formula and Demovire's Theorem.
• at about , the instructor states that (a^b)^d=a^bd. How is it possible for the exponent to shift down a degree?
• This is how exponents work. It doesn't shift it down a degree. Here are a few cases and examples:

(2^3)^2=8^2=64
(2^3)^2=2^(3*2)=2^6=64 (2^1=2,2^2=2,2^3=8,2^4=16,2^5=32,2^6=64)

(3^2)^3=9^3=729
(3^2)^3=3^(2*3)=3^6=729 (3,9,27,81,243,729)

This is not unlike what happens when you multiply two numbers with the same base raise to different powers. In that case you add the exponents:

2^2*2^4=2^(2+4)=2^6=64
2^2*2^4=4*16=64

Good question. It is confusing at first, but when you get the hang of it, it's not too hard.
• how would you find x in 36^x=6?
• Start by making both sides have a common base.
36 can be written as 6^2 and remember 6=6^1. This changes the equation into:
(6^2)^x = 6^1
Simplify: 6^(2x)=6^1
Now the only way for these 2 sides to be equal is for the exponents to be equal. So: 2x must equal 1
Solve 2x =1
x = 1/2

Hope this helps.
• does this rule apply to when we have the whole logarithm raised to a power, instead of just the argument? Ex:log_3(x^3)=3log_3(x), (log_3(x))^3=3log_3(x).
• In general, (log_b(a))^p will not equal p * log_b(a). The power rule with logarithms only applies when the logarithm input is raised to a power.
• Having a problem what is log(1000)= 4.605? This problem doesn’t seem to make sense to me and many of the problems following are similar
• log (1000) = 4.605 is an equation, not an expression. You can't solve an equation without any variables. Did you meant that the base of the logarithm was unknown?
• And yet another doubt I stumbled upon while exercising:
Give the domain of
f(x)=log(base3)(4x-3)^2

First solution:
2*log(base3)(4x-3)
4x-3>0
x>3/4

Second solution:
log(base3)(4x-3)^2
(4x-3)^2>0
x different than 3/4

Why is the first solution wrong?
• The domain depends on the positioning of the square, but your notation is ambiguous.
If you meant: f(x) = log₃[(4x-3)²]
Then the domain is x≠ ¾ because that value would lead to a log of 0. The square makes sure we never get a negative value for the argument.

If you mean f(x) = {log₃[(4x-3)}² which is usually written as log²₃ (4x-3) then:
As before x≠ ¾ due to being a log of 0. However, you are squaring the entire log, not just the argument. Thus, the argument can be negative, and the log of a negative number is not real, so that is outside the domain. Thus, the domain is
x > ¾

Again, please remember, the property log (aⁿ) = n log (a) ONLY APPLIES if a > 0. Therefore, you shouldn't use this property with variables that might lead to the argument being negative because that may lead to extraneous solutions.
• How would you solve 6 log x=-3 ?